$\begin{aligned} h(x)&=(5-6x)^5 \\\\ h'(x)&=\,? \end{aligned}$ Choose 1 answer: Choose 1 answer: (Choice A) A $-30(5-6x)^4$ (Choice B) B $(-6)^5$ (Choice C) C $5(5-6x)^4$ (Choice D) D $-6x^5+5x^4(5-6x)$
Explanation: Since $h$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $h(x)=\underbrace{(~\overbrace{5-6x}^{\text{inner}}~)^5}_{\text{outer}}$ So if $h(x)=w(u(x))$, then: $\begin{aligned} {u(x)}&={5-6x} &&\text{inner function} \\\\ w(x)&=x^5&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={-6} \\\\ {w'(x)}&={5x^4} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} h'(x)& =\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={5({5-6x})^4} \cdot {-6} \\\\ &=-30(5-6x)^4 \end{aligned}$